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Graph Theory, Part 1
1 The seven bridges of Konigsberg
1.1 The Problem
The city of Konigsberg (formerly in Prussia, now a part of Russia and called Kaliningrad) issplit by the River Pregel into various parts (including the island Kniephof), and back in theday there were seven bridges connecting the various parts, as you can see in the map below:
Koenigsberg
Kniephof
Pregel River
The city-folk wanted to know if it was possible to make a walking tour of the city, starting andending at the same place, in such a way that the walkers crossed each bridge exactly once. Thiswas considered a difficult problem at the time. After all, there are many different routes youmight consider, and it would be tedious to check all of them. Of course, if you find somethingthat works, you are done. But people couldn’t seem to make anything work, and so the questionarose whether there was not something inherent in the problem that made it insoluble.
In a 1736 paper which is considered the birth of a branch of math known as Graph Theory,the great mathematician Leonhard Euler showed decisively that it is impossible. His idea is sostunningly clear that it sets the whole matter to rest.
1.2 Abstraction
To help us deal with the problem more readily, let us abstract from the problem only the relevantinformation. Basically there are four pieces of land and various bridges connecting them. Wedon’t care what shops you visit or things you go to see on the way when you are not crossingbridges—for us each of the four pieces of land is just a place with bridges to some of the otherpieces of land. Let’s draw a dot for each of the four pieces of land. Visiting the each piece ofland is for us the same as visiting the dot in that land.
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Now let’s make connections between the dots, with one connection for each bridge:
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We can now remove all the burdensome geographical details:
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And draw the dots and connections more neatly. All we are concerned with is the connectivityof the lands by the bridges, not exactly how far or in what direction you need to walk:
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Now the problem is to go from dot to dot using each bridge only once. Suppose we starton the rightmost dot. Then when we leave it, we use one of the three bridges, and so there arestill two bridges attached to the rightmost dot which we eventually must use. The only way touse another is to return to that dot. When we do, we use another bridge, and are left with one
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bridge to use. Well then, we must leave that dot again, and use the last of the three bridgesattached to it. But then there is no bridge with which to come back, and we are supposed todo a round trip.
OK. Maybe we shouldn’t start the trip at the rightmost dot. We will start somewhere else.Nonetheless, we do need to visit the rightmost dot at some point in order to make use of bridgesattached to it. On our first visit, we use one bridge coming in. Then we must use another,and so we leave. But there is still one more bridge attached to the right dot that has not beenused. So we need to return again, but then we can’t leave! This is bad, because now our tripis a round trip which didn’t start at the rightmost dot.
In fact, it is not hard to show that every dot (piece of land), whether it is the one you startand end at or not, must have an even number of bridges attached, otherwise you will haveproblems. You use an even number of bridges when you come to a place and then leave again.For dots other than your starting dot, you must always leave after coming, so you always useup a pair of bridges per visit. So there must be an even number of bridges. For you startingpoint/destination dot, you use a pair of bridges for each passing visit, but also one bridge toleave at the beginning and one bridge to return at the very end. So it should also have an evennumber of bridges. In Konigsberg, every land has an odd number of bridges attached, so it ishopeless.
In honor of Euler, who showed that this cannot be done, a path that uses every bridge onceis called a Eulerian circuit. Konigsberg has no Eulerian circuit because it has regions with oddnumbers of bridges.
1.3 A Modification
OK. So we are definitely ruined unless every dot has an even number of bridges attached. Butmight we still have problems even if we do have an even number of bridges for every land?
For instance, suppose we build two more two bridges in Konigsberg so that each dot has aneven number of bridges:
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Now is it possible to make an Eulerian circuit? You can work out that it is indeed possible. Isthis a general rule for all graphs? It turns out it is.
Theorem 1 (Euler, Informal Version). It is possible to make an Eulerian circuit if andonly if all vertices (dots) have an even number of edges (bridges) [and the whole system isconnected—if one part is completely cut off from another, it is of course hopeless!].
We have proved that you cannot make an Eulerian circuit if you do not satisfy the “all-even” condition. How can you prove than you can make an Eulerian circuit if you do satisfy the“all-even” condition? The idea is as follows: start off wherever you want and walk over bridges
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as you please with the only rule being that you never use the same bridge twice. Eventuallyyou will get stuck since bridges keep getting used out. If you (by chance) managed to make anEulerian circuit, then you are done!
But maybe you didn’t. For instance, in the last graph above, you might have just startedat the top dot, walked to the left dot and then back again, and then walked to the right dotand back again. In this case, you would have used all four bridges from the top dot, namely:
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and you can no longer leave! But many bridges are left untraveled, so it is not an Euleriancircuit. What you do in this case is you go to your partly completed circuit, and pick a dotwhose bridges are not all used—for instance the right-hand dot. Then start another walk there,where you only use bridges that haven’t been used yet:
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AB
C
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After traversing A-B-C, we are stuck again at the rightmost dot, because we have used all thebridges from that dot. We combine the two tours, by going on the first one (1-2-3-4) until weget to the departure point of the second tour (the right-hand dot). Then we make the secondtour (A-B-C) a side trip, and after we finish the side trip, we complete the original tour. Inthis case, that means we start the original tour 1-2-3, then take the side trip A-B-C, and thenuse 4 to finish the original tour. So we now have a bigger tour 1-2-3-A-B-C-4, which we relabel1-2-3-4-5-6-7.
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There are still two bridges unused. between the left and bottom dot. So we make those twobridges into a little tour Y-Z, and then combine it with our existing tour 1-2-3-4-Y-Z-5-6-7.
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Each of the mini-trips is a circuit, so it uses an even number of bridges for each dot visited. Sowhat is left of the original graph at each stage after adding the side-trip is a smaller number ofunused bridges, with an even number of unused bridges for each dot (because an even number ofbridges minus an even number of bridges used is equal to an even number of bridges remaining).So we can always continue adding side-trips until all the bridges are used!
2 The Basics
2.1 What is a Graph?
The diagrams which we used to represent the Konigsberg problem economically are mathemat-ical objects called graphs. Euler’s paper on the Konigsberg problem is considered the birth ofthe branch of mathematics now known as Graph Theory.
Let’s make some definitions.
Definition 1 (Graph). A graph is an assembly of two kinds of things, vertices and edges. Theonly rule is that each edge starts at a vertex and ends a vertex.
We usually write vertices as dots and edges as lines between the two vertices they join. Wedo not need to draw the edges as straight lines, and in general we will not be able to avoiddrawing pictures where the edges displayed seem to touch each other. In fact, such edges shouldnot be considered to touch each other, but you can imagine that one goes over the other, muchlike a highway overpass. This is why we always draw the vertices as moderately-sized dots—sothat we are not confused into thinking that the crossed edges in
meet at some vertex. Since there is no vertex-dot, they do not. Note that in our Konigsbergproblem, we had more than one edge connecting a pair of vertices, to represent multiple bridgesthat accomplish the same thing. It is even possible to have an edge connecting a vertex toitself; this is called a loop. Here we have added a loop to the last graph:
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Sometimes it is important to assign directions to the edges on a graph, and so we define agraph with this additional information:
Definition 2 (Directed Graph). A directed graph is a graph where we have put arrows toassign directions to all the edges.
For example, the last graph which designates our Eulerian circuit in the modified Konigsbergbridge problem is a directed graph.
2.2 Degree
In the Konigsberg bridge problem, we were searching for a Eulerian circuit, which can now bedefined:
Definition 3 (Circuit, Eulerian Circuit). A circuit in a graph is a path which begins andends at the same vertex. A Eulerian circuit is a circuit in a graph which traverses each edgeprecisely once.
We found that a Eulerian circuit exists if and only if each vertex (representing a piece ofland) has an even number of edges (bridges) attached to it. We coin a new term to make iteasier to speak of such conditions:
Definition 4 (Degree). The degree of a vertex is the number of edges attached to that vertex.
For example, if we return to the Konigsberg bridges graph,
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then we see that the degrees of the vertices (starting at the top an proceeding clockwise) are3, 3, 3, and 5. Now we can state Euler’s result succinctly:
Theorem 2 (Euler, Formal Version). A graph has a Eulerian circuit if and only if allvertices have even degrees.
Thus the Konigsberg bridges graph has no Eulerian circuit.
2.3 Distance and Diameter
Definition 5 (Distance). The distance from vertex u to vertex v is the minimum number ofedges one must traverse to get from u to v.
In the undirected Konigsberg bridges graph vertex u is distance 1 from all other vertices,but vertex v is distance 2 from vertex x. We have a special name for vertices which are atdistance 1 from each other.
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Figure 1: (Undirected) Konigsberg Bridges Graph
Definition 6 (Adjacency). Two vertices in a graph are said to be adjacent if they are joinedby an edge.
If one vertex cannot be reached from another by a path made of edges, then we say thattheir distance from each other is ∞. If this happens, it means that the graph is not connected.For instance the graph
x
y
has the distance from vertex x to vertex y equal to ∞, and so the graph is not connected. Onthe other hand, the Konigsberg bridges graph (see Figure 1) is connected.
NOTE: We shall ALWAYS assume our graphs are connected unless explicitly noted otherwise.
When we label the vertices of a graph, as we have done in the graph of Figure 1, we alsoobtain a compact notation for labeling edges. For example, the edge connecting u to v is calleduv. Note that uv = vu, since this is an undirected graph, so that the order of the vertices doesnot matter.
In a directed graph, we only count paths that follow the one-way signs attached to the edges.For example, if we assign directions to edges in the graph of Figure 1:
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then the distance from w to x is 1, but the distance from x to w is 3. So distance is notsymmetric in a directed graph, as anyone knows who has driven in a city with lots of one-waystreets!
In a directed graph, we label edges in a way that indicates which way the arrow goes. Forexample, in the above graph, we have edge −→wx, also known as ←−xw, but we DO NOT have theedge −→xw =←−wx.
Definition 7 (Diameter). The diameter of a graph is the maximum of the distances betweenpairs of vertices.
The undirected Konigsberg bridges graph has diameter 2 (because every vertex is onlydistance 1 or 2 from the others), while an octagonal graph has diameter 4 (draw it and find thelongest distance).
2.4 Polygons, Girth, and Trees
The octagonal graph we just mentioned is one member of an important family of graphs. Letus consider undirected graphs for the time being.
Definition 8 (Polygon). A polygon is a sequence of edges and vertices where no edge orvertex is visited more than once, with the exception that the first vertex and the last coincide.[Note that Eulerian circuits need not be polygons: although they never re-use edges, they donot always meet the restriction on multiple visits to vertices.] The length of a polygon is thenumber or edges in the polygon.
The following graph has a polygon of length 6 in red.
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The graph also has polygons of lengths 5 and 7, but none of length less than 5, nor any oflength greater than 7. This motivates the next definition:
Definition 9 (Girth). The girth of a graph is the length of the shortest polygon in the graph.
A graph of girth 1 must have a loop. A graph of girth 2 must have no loops, but must havetwo vertices connected thus:
We call this strange object a 2-gon. A graph of girth 3 must have no 2-gons, nor loops, but musthave three vertices connected like a triangle (although they may be drawn with curvy edgesin our drawing of the graph). For the purposes of graph theory, we do call such a structure atriangle, however it may be drawn. For example, the following graph has a triangle in red:
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Once we allow this latitude in the use of the terms “triangle,” “square,” pentagon,” and so on,we see that a graph of girth 4 is free of loops, 2-gons, and triangles, but does contain a square.
What if the graph has no polygons at all? Then we define the girth to be ∞. We have aspecial name for such graphs.
Definition 10 (Tree). A tree is a graph containing no polygons.
So a tree has no loops nor 2-gons, so that every pair of vertices is either not joined orjoined by a single edge. Trees are widely used in such fields as biology, computer science, andlinguistics in organizing data. It is common to draw trees to look like upside-down versions ofthe trees you see in the real world:
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Here is a tree that represents the parsing of an English sentence:
the cat eats
the mouse
3 Isomorphism of Graphs
Two graphs G and H will be called isomorphic if they have the same pattern of connectivity.By this we mean that you can convert G into H by moving the vertices of G around withoutchanging which vertices the edges are connected to. You might also need to bend the edgesinto new shapes, but without changing which vertices they join. Don’t worry if you need edgesto move over other edges and vertices when making these changes.
For an example of where isomorphism is important, we turn to chemistry. Hydrocarbonsknown as alkanes may be represented by trees in which each vertex represents a carbon atom.The edges represent bonds between carbon atoms. A carbon can be bonded to at most fourother carbon atoms, so the maximum allowed degree of a vertex is 4. (The hydrogen atoms arenot depicted in these graphs, but a chemist automatically knows where to put them if shownthe graph of the carbon atoms.) The following two forms of octane (alkanes with 8 carbonatoms) are actually isomorphic:
We show the movements of vertices that make the first graph look like the second. We colorred the elements that are moved in each step. We should rotate the final figure by 180 degreesto finish the process.
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The following two graphs are not isomorphic, because one has a vertex of degree 4, whilethe other does not:
4 The Chinese Postman Problem
The Chinese Postman Problem is a generalization of the Konigsberg bridges problem. A Eu-lerian circuit exists only in graphs where all degrees are even. If not, we might ask ourselves,”What is the next closest thing to a Eulerian circuit?” This could be answered in several ways:the one we shall propose is adumbrated in the exploration where we constructed extra bridgesin Konigsberg in order to make all degrees even, thus enabling a Eulerian circuit to exist.
What we seek is a circuit that uses each edge once, and perhaps some edges more than once,but we would like to reuse edges as few times as possible. That is, we want a circuit employingall edges but of minimum total length (as measured in number of edges traversed). This is nota pointless quest: suppose we need to deliver mail, or sweep streets, or pick up garbage. Eachsegment of street between intersections could be represented as an edge, and the intersectionsas vertices. Then our insistence that each edge be traversed at least once is the insistence thatall houses in town receive services; retracing is tolerated as needed to make a full circuit, but nomore than necessary. Of course, in a real-world situation, we would note that different segmentsof streets are of different length, and we would try to minimize the total length (in meters) ofstreet segments retraced, rather than the total number of segments of streets retraced. This ismore complicated, so in these notes we shall stick to the simpler version, where we just try tominimize the number of edges in our circuit. In the homework, the more general problem willbe discussed. This more complicated problem is called the Chinese Postman Problem, afterthe Chinese mathematician Meigu Guan who studied it, and our easier version is called theSimplified Chinese Postman Problem.
Using an edge uv of a graph G twice in a circuit is much like having two copies of the edgeand using each copy once. In fact, let us make an extra copy of uv for each time we use itafter the first traversal. We then use the duplicates instead of the originals, so that no edge istraversed multiple times. And let us do the same for all other edges, add a duplicate for everyuse beyond the first and traverse duplicates so as to avoid double-use of any edge. What wesee is that a circuit in a graph G that traverses all edges at least once and which includes n
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additional traversals of edges already used is equivalent to a Eulerian circuit in a graph obtainedby adding n extra edges to G, where the added edges are copies of various edges existing in theoriginal graph G. Thus a minimum-length circuit that uses all edges in G is a Eulerian circuitin graph H obtained from G by duplicating as few edges as possible. We know that a Euleriancircuit will exist in H if and only if all degrees of vertices in H are of even degree. So our goalis to add duplicate edges to G—as few as possible—to make a graph H with all degrees even.This prompts a definition.
Definition 11 (Eulerization, Minimal Eulerization). A Eulerization of a graph G is agraph H, all of whose vertices are of even degree, obtained from G by duplicating edges. Aminimal Eulerization of G is a Eulerization of G with as few extra edges as possible.
So what we are looking for in the Simplified Chinese Postman Problem is a minimal Euler-ization of a given graph. For example, the Konigsberg bridges graph
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has all four vertices of odd degree. With each edge we add, we can at best hope to change twovertices to even degree. So we must add at least two edges. Thus, our modified Konigsbergbridges graph
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which duplicates two edges, is a minimal Eulerization of the original graph.Is it always possible to Eulerize a graph? Yes! A Eulerization that is probably not very
efficient is one that adds precisely one extra copy of each existing edge. This will definitelymake all degrees even. This also shows that the postman certainly does not need to more thandouble the length of the streets serviced.
5 Graphs as Models of Adjacency
Graphs are used as a basic mathematical model for showing adjacency or any other relationshipthat occurs between pairs of individuals. For example, one graph has all the world’s mathemati-cians as vertices. Two vertices (mathematicians) are joined by an edge if they are co-authorsof a joint paper. One particular mathematician, Paul Erdos, is famous for having co-writtenwith so many people, that most seasoned mathematicians are not far in distance from Erdosin the co-authorship graph. Here is a tiny portion of the authorship graph
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Erdos
Rota
Ulam
Metropolis
MullinKac
Graham
Chung
Odlyzko
The distance in the graph between a mathematical author and Erdos is given a name, “theErdos number.” Erdos has an Erdos number of zero, and people who coauthored with him arejoined to him with an edge in the graph, so they have Erdos number one. People who wrotewith someone who wrote with Erdos have Erdos number two. And so forth.
We could also graph the internet by making a site for each vertex. In our graph, an edgebetween sites indicates a link. But internet links have a direction—Site A can have a linksending you to Site B, but Site B may not have a link back to Site A. Thus, we should use adirected graph to represent the links on the internet.
Princeton Academics
Fine Hall Library
Princeton Math
Princeton Library Main
Princeton MainDepartments
Princeton Academic
5.1 Small World
When we start to look at large graphs like the coauthor graph or the internet graph, we oftenfind what is called the small world phenomenon. The first issue is one of connectedness. Sincepeople can make personal webpages that are unlinked to anything, we shall not always be ableto get from one page to another by following links. However, in these large graphs it is often thecase that there are many tiny isolated components, like little islands, along with one giganticconnected component, like a huge continent.
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The second issue deals with how closely connected things are in this giant component. Thisis characterized by the diameter of the graph—the maximum distance you need to go to getfrom one vertex to another. Around the year 2000, people estimated that there were about800 million pages in the web. However, researchers from Notre Dame performed an automatedsurvey of a portion of the internet and concluded that the diameter is about 19, i.e., any twopages in the giant connected component are 19 links apart or less.
6 Flows in Networks
One type of problem that graph theory has been used to model and address is the problem offlows in networks. A network could be a road system, and the flow could be the movement ofcars from one place to another on the roads. Or a network could be a computer network, withdata flowing from one computer to the next. Or a sewer system, or an electrical distributiongrid, and so forth.
6.1 The Model
We shall model our networks as directed graphs, and on each edge we shall write a number,which is the capacity of that edge. The capacity determines how much stuff (cars, data, sewage,energy) can flow along that edge in a fixed unit of time. Thus small roads or small pipes willhave low capacities, and big pipes or big roads will have large capacity numbers. We shallalways make our capacities integers. For now, we shall be interested in the somewhat restrictedproblem of moving all the stuff from one point (called the source) to another (called the target).We label the source s and the target t. For convenience, we shall always assume that no edgespoint into the source s or out of the target t. Here is an example of a network, with source,target, and capacities:
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2 2
3 t
4 10 7 9
9
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3
1
a
b
c
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e
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Suppose that this is a water distribution system, where the edges are pipes, the vertices arepumping stations, the direction of a given edge indicates which direction our pumps are set up
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to move water, and the capacities are the maximum number of cubic feet per second that canbe moved through the various pipes. Our goal will be to move as much stuff from s to t aspossible, with the following three constraints:
1. We may only move water through a pipe according to the direction it has.
2. We cannot move more water through a pipe than the capacity dictates.
3. The net flow into a pumping station must be exactly equal to the net flow out (otherwisewe would accumulate water or have a deficit, but we don’t have storage tanks in oursystem). The only stations exempt from this rule are the source s and the target t.
To respect these rules, we define a feasible flow to be an assignment of numbers to the edgessuch that each number is less than or equal to the capacity (to respect rule 2) and such that eachvertex is balanced, meaning the sum of the numbers that we have assigned to edges pointinginto the vertex is precisely the same as the sum of the numbers on the edges coming out of thevertex. This balance condition respects rules 1 and 3.
Here is an example of a feasible flow. We write our actual assignments in smaller type andin red—note that they are all less than or equal to the capacities, and check the balance of eachvertex (except s and t):
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2 2
3 t
4 10 7 9
9
3
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1
4
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7 7
1
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3
a
b
c
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7
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For an example of checking balance, look at the lowest vertex, which is labeled e. Edgespointing into it have assigned flows 7 and 7, so the total inflow is 14. Edges pointing out ofit have assigned flows 1, 5, and 8, so that the total outflow is also 14. Hence no water isaccumulating here.
Notice that the total flow out of the source is 3 + 5 + 4 = 12, and that the total flow intothe target is 2+2+8 = 12 also. No matter how tortuous the routing, it is inevitable that thesetwo numbers match since we do not let water accumulate at any other station.
Definition 12. The net flow from s to t in a feasible flow is equal to the sum of the flows onedges pointing out of s. In a feasible flow, this must be the same as the sum of the flows onedges pointing into t.
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6.2 Increasing Net Flow: Augmenting Paths
Our hope is to maximize the net flow from s to t. We note that the above assignment is notoptimal. Here is one way we could improve it:
1. We need to get more water out of the source. So increase the s→ c flow from 4 to 5.
2. Now we have an imbalance at c: an excess of one unit is incoming. We cannot increasethe outflow from c any more, since the edge from c to e is operating at full capacity. Soinstead we decrease how much water is going from e to c by 1. This restore balance to c.
3. But now e is unbalanced: an excess of one unit is incoming. So we increase the e → t
flow by 1, and this balances e.
4. Now one more unit of water is flowing into t (and one more unit flowing out of s). Wehave increased the net flow from t to s by one unit (from 12 to 13).
Here’s the new flow:
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4 10 7 9
9
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2
1
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5
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5 7
2
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7 7
1
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0
9
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This improvement we just made suggests the following definition:
Definition 13. Suppose we have a network with a feasible flow. An augmenting path is asequence E1, E2, . . . , En
of edges with the following properties:
1. If we ignore the directions of the edges, then E1, E2, . . . , Enform a path from s to t.
2. As we traverse this path, all forward-pointing edges have a flow that is less than fullcapacity, and all backward pointing-edges have a flow that is greater than zero.
The “path” (ignoring edge directions) −→sc,←−ce,−→et that we modified above is an example of an
augmenting path, since the forward-pointing edges (−→sc and−→et) were at less than full capacity,
and the backward-pointing edge (←−ce) had nonzero capacity.
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If we have an augmenting path, then increasing the total flow from s to t is straightforward.As we traverse the path, increase all flows on forward-pointing edges by 1 and decrease all flowson backward-pointing edges by 1. This is possible since the forward-pointing edges are not atfull capacity, and the backward pointing edges have positive flow. This change increases theflow out of s by 1, increases the flow into t by 1, and we claim that it retains balance at allother vertices. To see this, look in our example above: since we increased the flow on −→sc by oneand decreased the flow on←−ce by one, we have not changed the net flow into vertex c. Similarly,since we decreased the flow on ←−ce by one and increased the net flow on
−→et by one, we have
not changed the net flow out of vertex e. There are other possible combinations of increaseand decrease that we might see if we looked at other networks: generally speaking, we mightincrease the flow on both −→uv and −→vw, or decrease the flow both on ←−uv and ←−vw. Both of thesepreserve balance at v.
Let us return to our example, after we have increased the net flow from s to t to 13 by use ofthe augmenting path. Now we see that two of the pipes flowing into t are at maximum capacityand one is working at only one unit less than capacity. So perhaps we can improve the flowonce again. This would clearly maximize the net flow from s to t in the system since all pipesinto t would be at maximum capacity. We can try very hard to make incremental changes likebefore, but we seem to keep failing. But might there be a more clever way, perhaps involvinga radical change in the way we use the network, that will get us to a net flow of 14?
6.3 Obstructions to Improvements: Cuts
The answer is NO. The reason is as follows: let us cut our graph into two components, onecontaining s, b, c, and the other containing a, d, e, t. The edges within the first and secondcomponents are pink and green, respectively, but the edges between components are in blue.We only notate the capacities of the edges, and do not put any assigned flow.
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s 2 2
3 t
4 10 7 9
9
7
3
1
a
b
c
d
e
7
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Now note that stuff moves from the blue component to the green component fastest if all pipesleading from blue to green are at maximum capacity and all pipes going the opposite way are
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unused. Thus the maximum possible flow from blue to green is the sum of the capacities of−→sa,−→bd, −→ce, i.e., 3 + 3 + 7 = 13. Since nothing can get from s to t without crossing from the
blue component to the green component, we see that it would never be possible for the net flowfrom s to t to exceed 13. This PROVES that we cannot do better than a net flow of 13.
Now let us try to place the method we just used into a general context. We need to definewhat we did to the network above.
Definition 14. A cut of the network G is a partition of the vertices of G into two sets A andB, with s in set A, t in set B, and every other vertex is either in A or in B (but not both). Weidentify the cut with the ordered pair of sets (A, B). The capacity of (A, B) is the sum of thecapacities of all the edges that start at vertices in A and end at vertices in B.
Thus, in the paragraph above, our cut was the partition of the vertices {s, t, a, b, c, d, e} ofG into two sets: A = {s, b, c} and B = {a, d, e, t}. The capacity of the cut is found by addingthe capacities of the edges that start in A and end in B, i.e., we add the capacities of edges −→sa,−→bd, −→ce to get 3 + 3 + 7 = 13.
If we have a cut (A, B), then any flow from s to t must cross from A to B (since s is in A
and t is in B). It is clear that you cannot move more from A to B than the capacity of (A, B),so you cannot move more from s to t than the capacity of (A, B). Thus we have the followingprinciple:
The maximum net flow from s to t cannot be more than the minimum of the capacities of allcuts
That is, the lowest-capacity cut restricts the net flow from s to t to be less than or equal toits capacity. Actually, there may be multiple cuts that are equally low in capacity. These cutswhich are tied for the lowest capacity are called mincuts.
So the net flow from s to t cannot exceed the capacity of a mincut. Can it get that high?Or could it be that even this is not achievable?
6.4 Maximum Flow exactly determined by Mincut
In fact, one can achieve the capacity of a mincut—a challenge problem on the homework asksyou to prove that you can do this.
Theorem 3. In a network G, the maximum achievable net flow from s to t is precisely theminimum of the capacities of all the cuts of G.
Try to prove this in the homework!From all that we have learned up to this point, we now have a general strategy for maximizing
the net flow in a network:
1. Start with a flow that assigns zero flow to all edges. This is certainly feasible and balanced.
2. Find an augmenting path P from s to t.
3. Increase by one the flow of all forward-pointing edges on P , and decrease by one the flowon all backward-pointing edges of P .
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4. Repeat Step 3 until the path is no longer augmenting (i.e., some forward-pointing edge isworking at full capacity or some backward-pointing edge has zero flow).
5. Repeat Steps 2 to 4 until you cannot find any more augmenting paths.
6. Compute the net flow from s to t.
7. To prove that net flow from s to t is maximized, find a cut (A, B) of your network suchthat the capacity of (A, B) is equal to the net flow you have from s to t.
Let’s try this procedure with our network above. We start (step 1) by assigning zero flow toall edges.
6
s
3
2 2
3 t
4 10 7 9
9
7
3
1
a
b
c
d
e
70
0
0
0
0
0
0
0
0
0
0
0
0
0
Then we find an augmenting path from s to t (step 2). We have a lot of freedom here. I feellike using high-capacity pipes, so starting at s, I follow −→sc, then I am forced to use −→ce, the I
use−→ed because it is the biggest pipe out of e. It seems silly to go back to e again, so I use
−→dt.
(Not that this was really a very intelligent overall choice of path, but bear with me.)
6
s
3
2 2
3 t
4 10 7 9
9
7
3
1
a
b
c
d
e
70
0
0
0
0
0
0
0
0
0
0
0
0
0
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We increase the flow by one on each forward-pointing edge of the path (step 3). There are nobackward-pointing edges.
6
s
3
2 2
3 t
4 10 7 9
9
7
3
1
a
b
c
d
e
7
0
0
0
0
0
0
0
0 0
0
1
1
1
1
And then repeat this increasing two more times (step 4) until capacity limitations preventfurther increase.
6
s
3
2 2
3 t
4 10 7 9
9
7
3
1
a
b
c
d
e
7
0
0
0
0
0
0
0
0 0
0
3
3
3
3
Now step 5 tells us to go back to step 2 and find another augmenting path: now I decide to gowith −→sa and then
−→at.
6
s
3
2 2
3 t
4 10 7 9
9
7
3
1
a
b
c
d
e
7
0
0
0
0
0
0
0
0
0
0
3
3
3
3
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Both edges are forward-pointing; step 3 tells me to increase the flow on them by 1.
6
s
3
2 2
3 t
4 10 7 9
9
7
3
1
a
b
c
d
e
7
0
0
0
0
0
0
0 0
3
3
3
3
11
Step 4 tells us to repeat this procedure once more, at which point−→at cannot have its flow
increased any more, so that we no longer have an augmenting path.
6
s
3
2 2
3 t
4 10 7 9
9
7
3
1
a
b
c
d
e
7
0
0
0
0
0
0
0 0
3
3
3
3
22
21
Step 5 tells us to go back to step 2 and look for an augmenting path. I think I can make further
use of −→sa by proceeding to−→ad. I cannot use
−→dt, since it is working at full capacity. But I can
use←−de (which points toward me instead of away from me, but it is OK since it has nonzero
flow, so it can be part of an augmenting path). I complete the path by using−→et .
6
s
3
2 2
3 t
4 10 7 9
9
7
3
1
a
b
c
d
e
7
0
0
0
0
0
0
22
0
3
0
3
3
3
Now step 3 tells me to increase flows on the forward-pointing edges (−→sa,−→ad, and
−→et) by 1, while
decreasing the flow on the backward-pointing edge←−de by 1:
6
s
3
2 2
3 t
4 10 7 9
9
7
3
1
a
b
c
d
e
7
0
0
0
0
0
2
3
0
3
3
3 1
2
1
22
I can do no more with this path, since −→sa is at full capacity. So we look for yet another
augmenting path, and find one consisting of−→sb,−→bd,−→de, and
−→et .
6
s
3
2 2
3 t
4 10 7 9
9
7
1
a
b
c
d
e
7
0
0
0
0
0
2
3
0
3
3
3 1
2
1
3
Steps 3 and 4 tell us to increase the flow on these edges until one of them (−→bd) comes to full
capacity:
6
s
3
2 2
3 t
4 10 7 9
9
7
1
a
b
c
d
e
7
0
0
0
2
3
3
3
3 1
2
33
3
3 4
Then we look for another augmenting path: this time we try −→sc, −→ce,−→et .
6
s
3
2 2
3 t
4 10 7 9
9
7
1
a
b
c
d
e
7
0
0
0
2
3
3
3
3 1
2
33
3
3 4
23
Steps 3 and 4 tell us to increase the flow on these edges until one of them (−→sc, or, equally well,−→ce) comes to full capacity:
6
s
3
2 2
3 t
4 10 7 9
9
7
1
a
b
c
d
e
7
0
0
0
2
3
3 1
2
33
3
3
7
7
8
If we try to find another augmenting path, we need to use−→sb (since all other pipes out of s
are at full capacity). Then the only way to proceed is to use−→bc (since
−→bd is at full capacity
and←−bd has zero flow). Then the only way to proceed from there is to use ←−cs (since −→ce is at full
capacity and ←−ce has zero flow). But now we are back at s again, so there is no way to get to t.This makes us think that we have maximized the net flow.
We compute the net flow (step 6) to be 3 + 7 + 7 = 13. All we need to do is find a cut thathas capacity 13 (step 7) and we will have proved that the net flow cannot be increased anyfurther. But we know of such a cut: recall that the cut ({s, b, c}, {a, d, e, t})
6
s 2 2
3 t
4 10 7 9
9
7
3
1
a
b
c
d
e
7
3
has capacity 13.
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